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I am trying to prove any one of the following statements

  1. $((p \to r) \lor (q \to s)) \to ((p \to s) \lor (q \to r))$
  2. $(p \to q) \lor (q \to r)$
  3. $((p \to q) \to q) \to (p \lor q)$
  4. $(p \to q) \lor (q \to p)$

in the logical system 4CL. All of (1), (2) and (3) imply (4). All four of these are classical tautologies and are also valid in FOUR (Bellnap's 4-valued semantics). As such they should be provable in 4CL. I am not seeing a way to get any one of these. This system does not have LEM, contraposition, disjunctive syllogism or any form of explosion. What it does have is weakening (positive paradox), self distribution with all of the equivalent statements such as contraction, assertion, importing, permutation, conjunctive assertion, etc. It also has all of the DeMorgan properties, associativity and commutativity for both conjunction and disjunction. It has double negation introduction and elimination, and all of the basic distribution properties. It has transitivity, conjunction and distribution introduction, prefixing and suffixing. Finally, it has Peirce's Law $((p \to q) \to p) \to p$.

I feel that the key is Peirce's Law, but most proofs that I have seen using it, use some of the statements which are not valid in 4CL. Using it without those is not "clicking" for me. However, it is an axiom of 4CL and so far, I have not otherwise found anything that requires it. It has to be there for a reason, and I am thinking that statements such as (1) - (4) are why it is needed. So far, I haven't found an approach that works.

Any clues, ideas or proofs would be much appreciated. Please use infix notation when replying, statements like Cpq are just completely unreadable once they are non-trivial.

Thanks very much.

Michael Lee Finney
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  • Please be more precise. For instance, what do you mean by "it doesn't have contraposition"? You mean it's not an axiom, or you mean it's not derivable? Because if you add Peirce's Law to minimal implication, then you have classical implication. And, since you have double negation elimination, you usually have contraposition, unless you mean something peculiar by "contraposition". Perhaps you can cite a paper with a rigorous presentation of 4CL? – Hermógenes Oliveira Aug 20 '22 at 20:50
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    Yes, the paper is "Minimal Paradefinite Logics for Reasoning with Incompleteness and Inconsistency". You can find a copy with Google, but it is behind Springer's paywall. What you are saying about minimal logic is correct -- provided that you allow defining negation by $\bot \to p$.

    What I mean by "it doesn't have contraposition" is that it is neither an axiom nor can it be derived. That can be shown using Belnap's FOUR semantics.

     – Michael Lee Finney Aug 20 '22 at 23:27

1 Answers1

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I found a clue in the paper "Classifying Material Implications over Minimal Logic".

The way to use Peirce's Law to prove $P$, is to prove $(P \to Q) \to P)$. A convenient substitution is $P := p \lor (p \to q)$ and $Q := p$. Then detachment and Peirce's Law gives P.

Using that, I have proven 2 and 4. More, I have proven $p \lor (p \to q)$ from Peirce's Law and from that (2). Conversely, from (2) I have proven $p \lor (p \to q)$, and then Peirce's Law.

The forward direction did not require anything other than basic properties. No negation at all, no self distribution or anything related. Peirce's Law is only required in a single place (Ax 4). The properties that I used are:

Ax  1. p, p -> q >- q
Ax  2. p -> (q -> p)
Ax  4. ((p -> q) -> p) -> p
Ax  8. p -> p v q
Ax  9. q -> p v q
Ax 10. (p -> r) & (q -> r) -> (p v q -> r)

Th   1. p, q >- p & q
Th  14. p -> p
Th 285. (p -> q) & (q -> r) -> (p -> r)
Th 286. (p -> q) & (p -> r) -> (p -> q & r)

The proofs in the forward direction are shown below (all of the proofs have been validated by a theorem verification system). Some lines are very, very long, so local definitions have been used for readability.

Th 300. p v (p -> q)
   1. Df A.            A := p v (p -> q)
   2. Df B.            B := p -> q
   3. Df C.            C := p -> A
   4. Df D.            D := A -> q
   5. Ax 8.            C
   6. Ax 2.            C -> (D -> C)
   7. Ax 1.   {5, 6}   D -> C
   8. Th 14.           D -> D
   9. Th 1.   {7, 8}   (D -> C) & (D -> D)
  10. Th 286.          (D -> C) & (D -> D) -> (D -> C & D)
  11. Ax 1.   {9, 10}  D -> C & D
  12. Th 285.          C & D -> B
  13. Th 1.   {11, 12} (D -> C & D) & (C & D -> B)
  14. Th 285.          (D -> C & D) & (C & D -> B) -> (D -> B)
  15. Ax 1.   {13, 14} D -> B
  16. Ax 9.            B -> A
  17. Th 1.   {15, 16} (D -> B) & (B -> A)
  18. Th 285.          (D -> B) & (B -> A) -> (D -> A)
  19. Ax 1.   {17, 18} (D -> A)
  20. Ax 4.            (D -> A) -> A
  21. Ax 1.   {19, 20} A
  22. Df A.   {21}     p v (p -> q)

.

Th 301. (p -> q) v (q -> r)
  1. Df A.            A := (p -> q) v (q -> r)
  2. Ax 2.            q -> (p -> q)
  3. Ax 8.            (p -> q) -> A
  4. Th 1.   {2, 3}   (q -> (p -> q)) & ((p -> q) -> A)
  5. Th 285.          (q -> (p -> q)) & ((p -> q) -> A) -> (q -> A)
  6. Ax 1.   {4, 5}   q -> A
  7. Ax 9.            (q -> r) -> A
  8. Th 1.   {6, 7}   (q -> A) & ((q -> r) -> A)
  9. Ax 10.           (q -> A) & ((q -> r) -> A) -> (q v (q -> r) -> A)
 10. Ax 1.   {8, 9}   q v (q -> r) -> A
 11. Th 300.          q v (q -> r)
 12. Ax 1.   {11, 10} A
 13. Df A.   {12}     (p -> q) v (q -> r)

.

Th 302. (p -> q) v (q -> p)
   1. Th 301. (p -> q) v (q -> p)

The proofs in the reverse direction require assertion, which is equivalent to self distribution, and is valid in 4CL, plus two additional properties.

Th  17. p v p -> p
Th  69. (p -> q) & (r -> s) -> (p v r -> q v s)

Th 605. p & (p -> q) -> q

The proofs in the reverse direction follows.

Th 303. p v (p -> q)
   1. Df A.            A := (p -> p) -> p
   2. Df B.            B := p -> q
   3. Df C.            C := p -> p
   4. Th 14.           C
   5. Ax 2.            C -> (A -> C)
   6. Ax 1.   {4, 5}   A -> C
   7. Th 14.           A -> A
   8. Th 1.   {6, 7}   (A -> C) & (A -> A)
   9. Th 286.          (A -> C) & (A -> A) -> (A -> C & A)
  10. Ax 1.   {8, 9}   A -> C & A
  11. Th 605.          C & A -> p
  12. Th 1.   {10, 11} (A -> C & A) & (C & A -> p)
  13. Th 285.          (A -> C & A) & (C & A -> p) -> (A -> p)
  14. Ax 1.  {12, 13}  A -> p
  15. Th 14.           B -> B
  16. Th 1.  {14, 15}  (A -> p) & (B -> B)
  17. Th 69.           (A -> p) & (B -> B) -> (A v B -> p v B)
  18. Ax 1.  {16, 17}  A v B -> p v B
  19. Th 301.          A v B
  20. Ax 1.  {19, 18}  p v B
  21. Df B.  {20}      p v (p -> q)

.

Th 304. ((p -> q) -> p) -> p
    1. Df A.            A := (p -> q) -> p
    2. Df B.            B := p v (p -> q)
    3. Df C.            C := p v p
    4. Df D.            D := B -> C
    5. Df E.            E := p -> p
    6. Th 14.           E
    7. Ax 2.            E -> (A -> E)
    8. Ax 1.   {6, 7}   A -> E
    9. Th 14.           A -> A
   10. Th 1.   {8, 9}   (A -> E) & (A -> A)
   11. Th 286.          (A -> E) & (A -> A) -> (A -> E & A) 
   12. Ax 1.   {10, 11} A -> E & A
   13. Th 69.           E & A -> D
   14. Th 1.   {12, 13} (A -> E & A) & (E & A -> D)
   15. Th 285.          (A -> E & A) & (E & A -> D) -> (A -> D)
   16. Ax 1.   {14, 15} A -> D
   17. Th 303.          B
   18. Ax 2.            B -> (A -> B)
   19. Ax 1.   {17, 18} A -> B
   20. Th 1.   {19, 16} (A -> B) & (A -> D)
   21. Th 286.          (A -> B) & (A -> D) -> (A -> B & D)
   22. Ax 1.   {20, 21} A -> B & D
   23. Th 605.          B & D -> C
   24. Th 1.   {22, 23} (A -> B & D) & (B & D -> C)
   25. Th 285.          (A -> B & D) & (B & D -> C) -> (A -> C)
   26. Ax 1.   {24, 25} A -> C
   27. Th 17.           C -> p
   28. Th 1.   {26, 27} (A -> C) & (C -> p)
   29. Th 285.          (A -> C) & (C -> p) -> (A -> p)
   30. Ax 1.   {28, 29} A -> p
   31. Df A.   {30}     ((p -> q) -> p) -> p

I have not yet found a proof for (1) or (3). If I do so, I will update this answer. It would be interesting if anyone can find a way to eliminate assertion or its related theses in the reverse proofs.

Michael Lee Finney
  • 205